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V125BC [204]
3 years ago
11

A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts

from rest and is on a level road, what speed ???? will he be going after 25.0 m?
Physics
1 answer:
AleksAgata [21]3 years ago
4 0

Answer:

15.45 m/s

Explanation:

F = 1.2 x 10^3 N, Fk = 615 N, u = 0, s = 25 m

Let the speed after traveling 25 m is v.

mass = F / g = 1200 / 9.8 = 122.45 kg

Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N

acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2

Use third equation of motion

v^2 = u^2 + 2 a s

v^2 = 0 + 2 x 4.78 x 25 = 238.88

v = 15.45 m/s

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Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

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radius of rod = 5.75 cm

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E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

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When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.

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uysha [10]

Answer:

option (d)

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